145. Binary Tree Postorder Traversal

https://leetcode.com/problems/binary-tree-postorder-traversal/submissions/

Given the root of a binary tree, return the postorder traversal of its nodes' values.

Example 1:

Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:

Input: root = [1,2]
Output: [2,1]

Example 5:

Input: root = [1,null,2]
Output: [2,1]

Constraints:

• The number of the nodes in the tree is in the range [0, 100].

• -100 <= Node.val <= 100

Follow up:

Recursive solution is trivial, could you do it iteratively?

Solution

1. Recursive

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> values;
        if (root == nullptr) return values;
        postorderTraversal(values, root);
        return values;
    }
    void postorderTraversal(vector<int> &values, TreeNode* root) {
        if (root == nullptr) return;
        postorderTraversal(values, root->left);
        postorderTraversal(values, root->right);
        values.push_back(root->val);
    }
};

2. Iterative

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> values;
        if (root == nullptr) return values;
        stack<TreeNode*> nodes;
        stack<TreeNode*> temp;
        temp.push(root);
        while (!temp.empty()) {
            nodes.push(temp.top());
            temp.pop();
            if (nodes.top()->left) temp.push(nodes.top()->left);
            if (nodes.top()->right) temp.push(nodes.top()->right);
        }
        values.reserve(nodes.size());
        while (!nodes.empty()) {
            values.push_back(nodes.top()->val);
            nodes.pop();
        }
        return values;
    }
};

• #tree

• #bfs

• #dfs